Question: Proteins can be cleaved specifically at the amide bond on the carboxyl side of methionine residues by reaction with cyanogen bromide, BrC = N. The

Proteins can be cleaved specifically at the amide bond on the carboxyl side of methionine residues by reaction with cyanogen bromide, BrC = N. The reaction occurs in several steps:

(a) The first step is a nucleophilic substitution reaction of the sulfur on the methionine side chain with BrCN to give a cyano sulfonium ion, [R2SCN]+. Show the structure of the product, and propose a mechanism for the reaction.

(b) The second step is an internal SN2 reaction, with the carbonyl oxygen of the methionine residue displacing the positively charged sulfur leaving group and forming a five-membered ring product. Show the structure of the product and the mechanism of its formation.

(c) The third step is a hydrolysis reaction to split the peptide chain. The carboxyl group of the former methionine residue is now part of a lactone (cyclic ester) ring. Show the structure of the lactone product and the mechanism of its formation.

(d) The final step is a hydrolysis of the lactone to give the product shown Show the mechanism of thereaction.

|| 1. BrCN NHCHNHCH-C-OH + H2NHC- NHCHNHCH-C-NHCHC- R' 2. H20 CH2 '

|| 1. BrCN NHCHNHCH-C-OH + H2NHC- NHCHNHCH-C-NHCHC- R' 2. H20 CH2 ' CH2 CH2-S-CH3 CH2 O=U

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